Problem: Igor the analyst has adopted n little bunnies. As we all know, bunnies love carrots. Thus, Igor has bought a carrot to be shared between his bunnies. Igor wants to treat all the bunnies equally, and thus he wants to cut the carrot into n pieces of equal area. 

Formally, the carrot can be viewed as an isosceles triangle with base length equal to 1 and height equal to h. Igor wants to make n - 1 cuts parallel to the base to cut the carrot into n pieces. He wants to make sure that all n pieces have the same area. Can you help Igor determine where to cut the carrot so that each piece have equal area?

<image> Illustration to the first example.

Input

The first and only line of input contains two space-separated integers, n and h (2 ≤ n ≤ 1000, 1 ≤ h ≤ 105).

Output

The output should contain n - 1 real numbers x1, x2, ..., xn - 1. The number xi denotes that the i-th cut must be made xi units away from the apex of the carrot. In addition, 0 < x1 < x2 < ... < xn - 1 < h must hold. 

Your output will be considered correct if absolute or relative error of every number in your output doesn't exceed 10 - 6.

Formally, let your answer be a, and the jury's answer be b. Your answer is considered correct if <image>.

Examples

Input

3 2


Output

1.154700538379 1.632993161855


Input

2 100000


Output

70710.678118654752

Note

Definition of isosceles triangle: <https://en.wikipedia.org/wiki/Isosceles_triangle>.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 1010, H = 1e5 + 10;
int n, h;
int Write[20];
int read() {
  int d = 0, f = 1;
  char c = getchar();
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') d = (d << 3) + (d << 1) + c - 48, c = getchar();
  return d * f;
}
void write(int x) {
  int t = 0;
  if (x < 0) putchar('-'), x = -x;
  for (; x; x /= 10) Write[++t] = x % 10;
  if (!t) putchar('0');
  for (int i = t; i >= 1; i--) putchar((char)(Write[i] + 48));
}
void judge() {
  freopen(".in", "r", stdin);
  freopen(".out", "w", stdout);
}
int main() {
  n = read();
  h = read();
  n--;
  double s = h * 1.0 / 2 / (n + 1), last = 0;
  for (int i = 1; i <= n; i++) {
    double x = sqrt((s + last) * h * 2);
    printf("%.12lf ", x);
    last = x * x / h / 2;
  }
  return 0;
}